Question

question 10
write the equation of a line in standard form with a slope of \\(\frac{5}{4}\\) and a y-intercept of 17.
\\(\bigcirc\\) \\(5x - 4y = -68\\)
\\(\bigcirc\\) \\(5x + 4y = -68\\)
\\(\bigcirc\\) \\(-5x - 4y = 68\\)
\\(\bigcirc\\) \\(5x - 4y = 68\\)

Explanation:

Step1: Recall slope - intercept form

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. Given $m=\frac{5}{4}$ and $b = 17$, we have the equation $y=\frac{5}{4}x+17$.

Step2: Convert to standard form

The standard form of a line is $Ax + By=C$, where $A$, $B$, and $C$ are integers and $A\geq0$.
First, multiply each term in $y=\frac{5}{4}x + 17$ by 4 to get rid of the fraction:
$4y=5x + 68$
Then, subtract $5x$ from both sides:
$- 5x+4y = 68$
Multiply each term by - 1 to make $A>0$:
$5x-4y=-68$

Answer:

A. $5x - 4y=-68$